sobolev embedding theorem in the smooth metric measure space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:28:42Z http://mathoverflow.net/feeds/question/94598 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94598/sobolev-embedding-theorem-in-the-smooth-metric-measure-space sobolev embedding theorem in the smooth metric measure space mathsnail 2012-04-20T04:49:23Z 2012-04-20T09:44:49Z <p>we know the sobolev embedding theorem of Saloff-Coste </p> <p>$\Big(\int_B|F|^{2q}d\mu\Big)^{\frac1q}\le e^{C(1+\sqrt KR)}V^{-2/n}R^2\int_B\Big(|\nabla F|^2+R^{-2}F^2\Big)d\mu$</p> <p>wtih $Ric\ge-(n-1)K$, for all '$B$' of radius $R$ and volume $V$, $F\in C^{\infty}_0(B)$, $q=n/(n-2)$.</p> <p>My question is whether this inequality was established in the smooth metric measure space,i.e. $(M,g,e^{-f}d\mu)$ with Bakry-Emery Ricci curvature bouneded below $Ric_f=Ric+Hess f\ge-(n-1)K$?</p> <p>Thank you!</p> http://mathoverflow.net/questions/94598/sobolev-embedding-theorem-in-the-smooth-metric-measure-space/94614#94614 Answer by Robert Haslhofer for sobolev embedding theorem in the smooth metric measure space Robert Haslhofer 2012-04-20T09:44:49Z 2012-04-20T09:44:49Z <p>The Sobolev-inequality holds for general metric measure spaces satisfying CD(K,n), in particular for your smooth ones.</p> <p>See e.g. Theorem 21.15 in Villani's book <a href="http://math.univ-lyon1.fr/~villani/Cedrif/B07D.StFlour.pdf" rel="nofollow">http://math.univ-lyon1.fr/~villani/Cedrif/B07D.StFlour.pdf</a></p> <p>Also note that the $L^2$-version of the Sobolev-inequality follows from the $L^1$-version by inserting a suitable power of the function and using Hölder.</p>